non-symmetric positive definite matrix!?

Is symmetry a necessary condition for positive (or negative) definiteness?

If not:

It can be proved that if $\mathbf{A}:(m\times m)$ is a square (non-symmetric) matrix, then $$\mathbf{z'Az=z'Bz},~~\mathbf{B=B'=(A+A')/2}$$

On the other hand, a positive definite matrix is a symmetric matrix for which:

$$\mathbf{z'Bz}>0,~~ \mathbf{z\ne o}$$

Can we imply that $\mathbf{A}$ which is a non-symmetric matrix, is positive definite?

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It looks like you are working with real matrices.

Most often the definition os positive definite includes symmetric, but sometimes this is not required.

In any case, if $z'Az\geq0$ then $z'A'z=(z'Az)'=z'Az\geq0$. So $$z'Az=\frac12(z'Az+z'A'z)=z'\left(\frac{A+A'}2\right)z.$$

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So, This means that if $\mathbf{z'Az}\geq0$, whether $A = A'$ or not, we call $\mathbf{A}$ a Positive definite Matrix? – Ramin Dec 9 '12 at 7:48
I wouldn't do it, but many people does. – Martin Argerami Dec 9 '12 at 10:30
You mean I can find a reference that defines Positive definiteness for non-symmetric matrices? – Ramin Dec 9 '12 at 14:00
The thing is that when you work with a complex vector space, you get Hermitian for free; but not in the real case. This issue with the definition is mentioned in the wikipedia article: en.wikipedia.org/wiki/… but I'm not familiar with references dealing with positive definite matrices. – Martin Argerami Dec 9 '12 at 14:48

Yes, it is possible. In fact, it follows easily from the properties of taking the transpose:

$$0 < z^{\mathrm{T}}\left(A+A^{\mathrm{T}}\right) z = z^{\mathrm{T}}Az + z^{\mathrm{T}}A^{\mathrm{T}}z = z^{\mathrm{T}}Az + (z^{\mathrm{T}}Az)^{\mathrm{T}} = 2z^{\mathrm{T}}Az$$

Taking the transpose of a real number doesn't change anything. Therefore $z^{\mathrm{T}}Az> 0$ if and only if $z^{\mathrm{T}}Bz>0$.

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Would you please see my comment at Martin Argerami answer. Thanks – Ramin Dec 9 '12 at 7:50